# Limit proof

1. Nov 3, 2006

### kreil

Show $$\lim_{x{\rightarrow}c}\sqrt{x}=\sqrt{c}$$

In other words, $$\forall \epsilon > 0 \exists \delta > 0 s.t. |x-c|< \delta \implies |\sqrt{x}-\sqrt{c}|< \epsilon$$

So $$0<|x-c|<\delta \implies |\sqrt{x}-\sqrt{c}|=|\frac{x-c}{\sqrt{x}+\sqrt{c}}|$$

and this is where im stuck, a hint the teacher gave was "bound the denominator" but im not sure how to do that...can anyone help?

Josh

2. Nov 3, 2006

### StatusX

You want to show:

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c} } < \epsilon$$

The way these things work is that you want to find something bigger than the LHS that is still smaller than RHS, because then the LHS is definitely smaller than the RHS. So keep replacing the LHS by something bigger and bigger until it's in a form that is easy to compare to the RHS. What can you replace the denominator by (that's independent of x) to get something larger than the LHS?

3. Nov 3, 2006

### kreil

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{1}<\delta=\epsilon$$

Can I do that and just choose delta = epsilon?

4. Nov 3, 2006

### StatusX

No, that isn't true. If x=c=1/16, then the denominator is 1/2, and so the LHS is not less than the middle. The bound will depend on c.

5. Nov 3, 2006

### kreil

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{3\sqrt{c}}$$

then this should work since if x=c=1/16, the LHS is 1/2 and the RHS is 3/4

can I then continue this so that

$$\frac{|x-c|}{3\sqrt{c}}<3\sqrt{c}\delta=\epsilon$$ .....??

6. Nov 3, 2006

### StatusX

No. First of all, you can't show something works with a single example. But it doesn't even work for that one. Remember, if you want to increase a fraction, you have to decrease its denominator. So you have to find something that is less than $\sqrt{x}+\sqrt{c}$.

7. Nov 3, 2006

### kreil

so how about just squareroot of c?

8. Nov 3, 2006

### StatusX

Yea, that's less than $\sqrt{x}+\sqrt{c}$. So what's your answer?

9. Nov 3, 2006

### kreil

In other words, choose $$\delta=\frac{\epsilon}{\sqrt{c}}$$

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{\sqrt{c}}<\sqrt{c} \delta = \epsilon$$

Last edited: Nov 3, 2006
10. Nov 3, 2006

### StatusX

Looks good.

11. Nov 4, 2006

### kreil

Another limit question

I need to show that $$\lim_{x{\rightarrow}a}x^3=a^3$$

then $$0<|x-a|<\delta \implies |x^3-a^3|< \epsilon$$

I'm having a lot of trouble getting the cubed terms to be less than an expression with |x-a|. Should I try turning the cubed terms into a fraction? I need a direction to go in.

Josh

12. Nov 5, 2006

### Muzza

x^3 - a^3 = (x - a)(x^2 + ax + a^2).

13. Nov 5, 2006

### 0rthodontist

Also you could use |x|, |a| < |a| + e. That turns the second factor into something less than 3(a + e)^2

14. Nov 5, 2006

### kreil

Yet Another E-d proof

I got that one now, thanks.

Now this one is giving me a problem...

Show $$\lim_{x{\rightarrow}1}\frac{x^2-x+1}{x+1}=\frac{1}{2}$$

i.e. show $$\forall \epsilon > 0 \exists \delta > 0 s.t. |x-1|< \delta \implies |\frac{x^2-x+1}{x+1}-\frac{1}{2}|<\epsilon$$

The teacher says to let $$x=1+ \lambda$$ so that as $$x \rightarrow 1, \lambda \rightarrow 0$$. I wasn't quite sure how to properly incorporate this into the proof, because it gave me problems choosing delta:

$$|\frac{x^2-x+1}{x+1}-\frac{1}{2}|=|\frac{2x^2-3x+1}{2(x+1)} |=|\frac{(1+ \lambda -1)(2+2 \lambda -1)}{2(1+ \lambda +1)}|=|\frac{(\lambda)(2 \lambda +1)}{2( \lambda + 2)}=0< \epsilon$$

I know that this can't be right since it doesnt even require me to choose a delta, so it doesn't really satisfy the definition. Any help incorporating the hint will be helpful.

Last edited: Nov 5, 2006
15. Nov 5, 2006

### 0rthodontist

lambda is not equal to 0, lambda is some constant. You want to show that when 0 < |lambda| < delta for some delta, your expression on the right is less than epsilon.